Diodes TEC 284

Diodes Conduct electricity in one direction only Silicon and Germanium are the main semiconductor materials used in manufacturing diodes, transistors and integrated circuits Semiconductor material is refined to make it pure Minute controlled amounts of impurity are then added in a process called doping

Silicon Crystal Silicon crystal in its pure form is an insulator No free electrons It has a lattice structure Each atom forms four covalent bonds with neighboring atoms

Diodes Two Dopant Types N-type (Negative) – Free flowing electrons are added to the silicon structure P-type (Positive) – Lack of electrons creates holes or slots which allow spaces for electrons to migrate to

Doping Silicon – N type Addition of impurities such as antimony, arsenic or phosphorous adds free electrons and greatly increases the conductivity of silicon This is an N type semiconductor region due to the excess of electrons Source: http://hyperphysics.phy-astr.gsu.edu/hbase/solids/dope.html#c3

Doping Silicon – P type Addition of impurities such as boron, aluminum or gallium creates deficiencies of electrons or “holes” This is an P type semiconductor region due to the deficiency of electrons Source: http://hyperphysics.phy-astr.gsu.edu/hbase/solids/dope.html#c4

PN junction When a semiconductor chip contains a N doped region adjacent to a P doped region a diode junction or PN junction is formed Junctions can be made from silicon or germanium, but the two are not mixed when making PN junctions In a PN junction P material is called the anode and N material is called the cathode

PN Junction Electric current will flow through a PN junction in one direction only

Forward Biased Diode When diode is connected so that the current is flowing it is said to be forward biased For forward biased diodes, anode is connected to a higher voltage than the cathode

Reverse Biased Diode When diode is connected so that the current is not flowing it is said to be reverse biased For reverse biased diodes, cathode is connected to a higher voltage than the anode The diode cannot conduct

Diode Assumptions In many circuits, the diode is considered to be a perfect diode to simplify calculations A perfect diode means a zero voltage drop in the forward direction and no current conduction in the reverse direction A forward based diode can be compared to a closed switch A reversed based diode can be compared to an open switch

I-V curve for a diode I-V characteristics for a P-N junction diode Image Source : http://en.wikipedia.org/wiki/Diode

Knee voltage For diodes there is a range or region where the diode resistance changes from high to low This is called the knee region The voltage at which the diode “turns on” is called the knee voltage For most silicon diodes, this is about 0.7 V For germanium diodes, it is approx 0.3 V

Knee voltage

Assumptions when using imperfect diodes The voltage drop across the diode is either 0.7 V or 0.3 V. In some instances when voltage are large, it is assumed that diodes are perfect and when conducting, the voltage drop across the diode is oV. Diodes are assumed to be perfect when voltage supply is 10V or more Excessive current is prevented from flowing through the diode by using a resistor in series with the diode

Calculation Calculate the current flowing through the diode below. Assume VD = 0.7 V

Answer I = (VS – VD) / VR I = (5 – o.7) / 1000 = 4.3 mA This is a series curcuit I is the same everywhere in the circuit

Power dissipation in diodes When current flows through a diode there is heating and power dissipation just like in a resistor Voltage drop for a silicon resistor is assumed to be 0.7 V For a silicon diode with 100 mA flowing through it P = IV = .1 x .7 = 70 mW

Question If a silicon diode has a max power rating of 2 Watts, how much current can it safely pass? P = IV I = P / V I = 2 /0.7 = 2.86A

Finding the current through a diode Steps Find I2 Find VR Find IT Find ID 5 V ID Find ID in the circuit above (the current through the diode)

Finding the current through a diode Steps Find I2 Find VR Find IT Find ID 5 V ID I2 = (VD / R2) = 0.7/70 = 10 mA VR = (5 – VD) = (5 – 0.7) = 4.3 V IT = VR / R1 = 4.3 / 43 = 0.1 A = 100 mA ID = IT – I2 = 100 – 10 = 90 mA

Diode Breakdown If you place the diode in the circuit backwards (reverse biased) then almost no current flows.

Diode Breakdown The I-V curve for a perfect diode would be zero current for all values For a real diode, when a certain voltage is reached, the diode “breaks down” and allows a large amount of current to flow If this is allowed to continue, the diode will burn out You can avoid burnout by limiting the current with a resistor

Diode Breakdown Breakdown is not catastrophic and does not destroy the diode The diode will recover and operate normally provided the current is limited to prevent burn out Breakdown Voltage is called the Peak Inverse Voltage (PIV) or Peak Reverse Voltage (PRV) Breakdown voltage varies from one type of diode to another

Zener Diode Diodes can be manufactured so that breakdown occurs at lower and more precise voltages These are called Zener diodes At the Zener voltage a small current will flow through the diode The current must be maintained to the keep the diode at the zener point

Zener Diode Zeners are used to maintain constant voltage at some point in a circuit

Zener Applications Lamp requires 20 V and 1.5 A Power source is a generator at 50 V Generator voltage may fluctuate causing lamp to get brighter and dimmer. Resistor calculation R is no longer valid for fluctuations. This may be unacceptable behavior

Zener Applications Alternative arrangement Zener diode used to maintain constant voltage across the lamp

Zener Considerations Choose values of resistors that would prevent the Zener diode from burning out This max current rating of the zener diode is given in the specs or can be calculated P = I V e.g A zener diode rated at 20 W which has a breakdown voltage of 10 V will allow at max 2 A to flow through it

Zener Calculation Find the current going through the Zener diode above IZ Find the current going through the Zener diode above

Zener Calculation IR = (50 – 20)/ 7.5 = 4 A IZ = IR – IL

Zener Calculation What happens if the input voltage changes to 35 V? IR = (35 – 20)/ 7.5 = 2 A IZ = IR – IL IZ = 2 – 1.5 = 0.5 A IR IZ

Zener behavior When the input voltage changes from 50 V to 35 V (as in the case when the voltage on a generator fluctuates), all other values in the circuit remains the same Only the current flowing through the zener diode changes When determining the value of R for your circuit, start off with the worst case possible voltage (in this case 35 V) which allows 1.5 A to flow through the lamp